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0

following Tito Piezas III comment I'll provide some concrete examples. First, let's take the second convergent of the cfrac $$G(q,k)\approx\cfrac{1}{1-q+\cfrac{1}{1-{q^3}^k}}$$ and expand it into a power series as $k\to\infty$ $$G(q,k)\approx \frac{1}{2}+\frac{1}{2^2}q+\frac{1}{2^3}q^2+\dots$$ Which converges to $\frac{2}{3}$ for $q=\frac{1}{2}$ as shown ...


2

Take $k=1$ and $q=1/2.$ Then the first few convergents are (starting with the trivial zeroth which here is $0$) $$C(0)=0,\\ C(1)=2,\\ C(2)=14/23=0.60869..,\\ C(3)=946/969=0.97626..,\\ C(5)=177486/217271=0.81688.$$ Since the terms $b(k)=1-(1/2)^{2k-1}$ go so rapidly to $1$ it would seem odd (to me) if the convergents were not alternately below/above the ...


1

The difference is that Alpha is giving you a negative continued fraction-note that it is $- [1; 1, 8, 1, 3, 2, 2, 9, \dots]$ Your answer key is giving you a positive continued fraction that is added to $-2$. If you add $3106$ to the numerator to make the value positive, Alpha gives $[0; 9, 1, 3, \overline{2, 2, 9, 1, 1, 2, 1, 4, 13, 9, 1, 9, 1, 118, 1, 9, ...


5

The answer is yes. Given the nome $q = \exp(i\pi\tau)$, elliptic lambda function $\lambda(\tau)$, Dedekind eta function $\eta(\tau)$, Jacobi theta functions $\vartheta_n(0,q)$, and Ramanujan's octic cfrac, the following relations are known, $$\begin{aligned} u(\tau) & = \big(\lambda(\tau)\big)^{1/8} = \frac{\sqrt{2}\, \eta(\tfrac{\tau}{2})\, ...


2

So with formal manipulation: $$\psi(1/q)= \frac{-q^{-1}}{1-q^{-1}+ \frac{q^{-1}(1-q^{-1})^2}{1-q^{-3}+\ldots}}$$ Times top and bottom by $-q$ to give: $$\psi(1/q)= \frac{1}{1-q+ \frac{-qq^{-3}(1-q)^2}{1-q^{-3}+\frac{q^{-1}(1-q^{-2})^2}{1-q^{-5}+\ldots}}}$$ Then mutiply top and bottom of the next fraction by $-q^3$ to give: $$\psi(1/q)= \frac{1}{1-q+ ...


2

To clarify, what you found is a q-continued fraction for the Jacobi theta function $\vartheta_2(0,q)$. Using ccorn's formulation, $$\left(\frac{\vartheta_2(0,q)}{2\,q^{1/4}}\right)^2 =\Big(\sum_{n=0}^\infty q^{n(n+1)}\Big)^2 ...


1

Regarding R. Israel's remark, three of your continued fractions, while not exactly the same, are variants of a common form discussed here for $|q|<1$, $$\frac{1}{1-q} =\cfrac{1}{1+q-\cfrac{\color{brown}{2q(1+q^2)}}{1+q^3+\cfrac{q^2(1-q)(1-q^3)}{1+q^5-\cfrac{q^3(1+q^2)(1+q^4)}{1+q^7+\cfrac{q^4(1-q^3)(1-q^5)}{1+q^9-\ddots}}}}}\tag0$$ First, the one for ...


1

This supplements R. Israel's answer. Given the continued fraction discussed in this post for $|q|<1$, $$\frac{1}{1-q} =\cfrac{1}{1+q-\cfrac{\color{brown}{2q(1+q^2)}}{1+q^3+\cfrac{q^2(1-q)(1-q^3)}{1+q^5-\cfrac{q^3(1+q^2)(1+q^4)}{1+q^7+\cfrac{q^4(1-q^3)(1-q^5)}{1+q^9-\ddots}}}}}$$ and using a little algebraic manipulation to transform the brown part to ...


2

(A partial answer.) This is a special case of a conjectured equality discussed in this MO post. Let $|q|<1$, then, $$\begin{aligned}U(q) &= \prod_{n=0}^\infty \frac{\big(1-a^2q^3(q^4)^n\big)\big(1-b^2q^3(q^4)^n\big)}{\big(1-a^2q(q^4)^n\big)\big(1-b^2q(q^4)^n\big)}\\ &= \dfrac{1} {1+ab-\dfrac{(a+bq)(b+aq)} {1+(ab)^3+\dfrac{(a-bq^2)(b-aq^2)q} ...


3

The ordinary generating function for your recurrence is $$ g(x) = \dfrac{1+x^2}{1-x-x^3}$$ Thus $$\sum_{n=0}^\infty (-1)^n a_n q^n = g(-q) = \frac{1+q^2}{1+q+q^3}\tag1$$ If that is $\phi(q)$, then indeed $$ \phi(1/q) = \dfrac{1/q^2+1}{1/q^3 + 1/q + 1} = \dfrac{q (1 + q^2)}{1+q^2 + q^3} = q \phi(q)$$ Now let's try to get your continued fraction. $$ ...


2

Let $F(x)=x^0+\cfrac1{x^1+\cfrac1\cdots}~.$ Then $F(2)$ is OEIS A$214070$, for which no closed form is currently known.


1

You want the value which is greater than $1$ - clearly by estimating. From $x^8+\frac 1{x^8}=2207$ you know that if $x$ satisfies that equation so will $\frac 1x$. Put $\frac 1x$ in the original equation and modify the eighth root accordingly and you will see where the alternative answer comes from and that this is clearly less than $1$.


3

The 8th root is not pertinent. You are asking how to assign a value to the continued fraction inside the 8th root. As the equation for $x^8$ shows, there are two solutions of $f = 2207 - \frac{1}{f}$. My first thought was that the even and odd "levels" of the continued fraction converge to the two different solutions (making it hard to distinguish the ...


0

Note: OP's expression is usually regarded as continued fraction. We will show it has the single solution $2$. Before we analyse OPs expression, let's have a look at a continued fraction representation of $\sqrt{2}$ \begin{align*} \sqrt{2}=[1;2,2,2,\ldots]=1+\frac{1}{2+\frac{1}{2+\frac{1}{2+\ddots}}} \end{align*} The convenient notation ...


2

Note that as we add more terms to the continued fraction, it oscillates between $1$ and slightly higher than $2$. $$ \begin{align} n&=1& 3&=3& 3-2&=1\\\\ n&=2& 3-\cfrac23&=\frac73& 3-\cfrac{2}{3-2}&=1\\\\ n&=3& 3-\cfrac{2}{3-\cfrac23}&=\frac{15}7& 3-\cfrac2{3-\cfrac2{3-2}}&=1\\\\ n&=4& ...


5

Let us define two series. The first is \begin{align} a_1 &= 3 \\ a_2 &= 3 - \frac{2}{3} \\ a_3 &= 3 - \frac{2}{3 - \frac{2}{3}} \\ a_4 &= 3- \frac{2}{3 - \frac{2}{3 - \frac{2}{3}}} \\ &\vdots \\ a_{n+1} &= 3 - \frac{2}{a_n} \quad (*) \end{align} and \begin{align} b_1 &= 3 - 2 \\ b_2 &= 3 - \frac{2}{3-2} \\ b_3 &= 3 - ...


6

This continued fraction is the limit of the sequence $a_n=3-2/a_{n-1}$. Computing the first few terms shows that $2$ is the correct limit; if our initial term $a_1=3$ were different then the limit could be $1$.


-1

Your two solutions are correct. You can verify them as follows. $$\color{blue}{1} = 3 - 2=3-\frac{2}{\color{\red}{1}}\tag{1}$$ Now replace the red 1 with the blue 1, which equals to the right hand side in (1). $$\color{blue}{2} = 3 - 1=3-\frac{\color{cyan}{2}}{\color{\red}{2}}\tag{2}$$ Now replace the red 2 with the blue 2, which equals to the right hand ...


2

Look at it as two different series and you will understand why both 1 and 2 are possible solutions of this: series 1: $\{3-2, 3-\frac{2}{3-2}, 3-\frac{2}{3-\frac{2}{3-2}}, ...\}$ series 2: $\{3-\frac{2}{3}, 3-\frac{2}{3-\frac{2}{3}}, 3-\frac{2}{3-\frac{2}{3-\frac{2}{3}}}, ...\}$. series 1 converges to 1 whereas series 2 converges to 2.


6

If you continue adding numbers to the expression one at a time, then you have the sequence $$ 3,\; 3-2,\; 3-\frac{2}{3},\; 3-\frac{2}{3-2},\; 3-\frac{2}{3-\frac{2}{3}},\; 3-\frac{2}{3-\frac{2}{3-2}},\;3-\frac{2}{3-\frac{2}{3-\frac{2}{3}}}\ldots, $$ or $$ 3,\; 1,\; \frac{7}{3},\; 1,\; \frac{15}{7},\; 1,\;\frac{31}{15},\;\ldots, $$ which consists of two ...


1

Also, if we take the house numbers $$ x_0 = 0, $$ $$ x_1 = 1, $$ $$ x_2 = 6, $$ $$ x_3 = 35, $$ $$ x_4 = 204, $$ $$ x_5 = 1189, $$ we have a simple recurrence $$ \color{magenta}{ x_{n+2} = 6 x_{n+1} - x_n }$$ which follows by Cayley-Hamilton from the generator of the oriented automorphism group of $u^2 - 8 v^2.$ If $y_n$ is the number of houses on the ...


2

If it is house number $x$ in a street of $y$ houses, we have $$\frac{x(x-1)}{2}+x+\frac{x(x-1)}{2}=\frac{y(y+1)}{2}$$ which simplifes to $$(2y+1)^2-8x^2=1\ .$$ This can be solved by computing the continued fraction $$\sqrt8=2+\frac{1}{1+{}}\frac{1}{4+}\frac{1}{1+{}}\frac{1}{4+\cdots}\ .$$ The table of convergents is ...


4

I will stick to verifying the identity numerically. A backwards recursion formula for the $n$'th partial quotient of the continued fraction is $$s_{k-1} = 1 - e^{-(2k-1)\pi} + \frac{e^{-k\pi}(1+e^{-k\pi})^2}{s_{k}}$$ for $k=n,n-1,\ldots,3,2,1$ and $s_n = 1$. Having calculated $s_0$ the $n$'th partial quotient is then given as $1 + \frac{2e^{-\pi/2}}{s_0}$. ...



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